Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
void move(int *x, int i, int j, int &nret, int ret[1000000][2]) {
  ret[nret][0] = i, ret[nret][1] = j, ++nret, x[i] ^= x[j];
}
void swap(int *x, int i, int j, int &nret, int ret[1000000][2]) {
  move(x, i, j, nret, ret);
  move(x, j, i, nret, ret);
  move(x, i, j, nret, ret);
}
void gaussjord(int n, int *x, int &nret, int ret[1000000][2]) {
  for (int i = (0); i < (n); ++i) {
    {
      int best = i;
      for (int j = (i + 1); j < (n); ++j)
        if (x[j] > x[best]) best = j;
      if (best != i) swap(x, i, best, nret, ret);
    }
    for (int j = (i + 1); j < (n); ++j)
      if ((x[i] ^ x[j]) < x[j]) move(x, j, i, nret, ret);
  }
  for (int i = n - 1; i >= 0; --i)
    if (x[i] != 0) {
      for (int j = (0); j < (i); ++j)
        if ((x[i] ^ x[j]) < x[j]) move(x, j, i, nret, ret);
    }
}
int n;
int s[10000];
int t[10000];
int ret1[1000000][2], nret1;
int ret2[1000000][2], nret2;
int ret[1000000][2], nret;
void run() {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) scanf("%d", &s[i]);
  for (int i = (0); i < (n); ++i) scanf("%d", &t[i]);
  nret1 = nret2 = 0;
  gaussjord(n, s, nret1, ret1);
  gaussjord(n, t, nret2, ret2);
  for (int i = (0); i < (n); ++i) {
    for (int j = (i); j < (n); ++j) {
      if (s[j] == 0) break;
      bool have = (s[i] ^ s[j]) < s[i];
      bool want = (t[i] ^ s[j]) < t[i];
      if (have != want) move(s, i, j, nret1, ret1);
    }
  }
  for (int i = (0); i < (n); ++i)
    if (s[i] != t[i]) {
      printf("-1\n");
      return;
    }
  printf("%d\n", nret1 + nret2);
  for (int i = (0); i < (nret1); ++i)
    printf("%d %d\n", ret1[i][0] + 1, ret1[i][1] + 1);
  for (int i = nret2 - 1; i >= 0; --i)
    printf("%d %d\n", ret2[i][0] + 1, ret2[i][1] + 1);
}
int main() {
  run();
  return 0;
}